Question:

From $p(x) = 3x^2 - 6x + 4$: $\alpha + \beta = -\dfrac{-6}{3} = 2$ and $\alpha \beta = \dfrac{4}{3}$. [1.0 Mark]
Expression $= \dfrac{\alpha^2 + \beta^2}{\alpha \beta} + 2\left(\dfrac{\alpha + \beta}{\alpha \beta}\right) + 3\alpha \beta = \dfrac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta} + \dfrac{2(\alpha + \beta)}{\alpha \beta} + 3\alpha \beta$. [1.0 Mark]
Substituting values:
$= \dfrac{2^2 - 2(4/3)}{4/3} + \dfrac{2(2)}{4/3} + 3\left(\dfrac{4}{3}\right) = \dfrac{4 - 8/3}{4/3} + \dfrac{4}{4/3} + 4 = \dfrac{4/3}{4/3} + 3 + 4 = 1 + 3 + 4 = 8$. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Finding $\alpha+\beta = 2$ and $\alpha\beta = 4/3$: 1.0 Mark
Simplifying expression into sum and product terms: 1.0 Mark
Correct substitution and evaluation to get 8: 1.0 Mark